In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$

Solution 1:

We can prove $$\cos A+\cos B+\cos C=1+4\sin\frac A2\sin\frac B2\sin\frac C2$$

Now, $0<\dfrac A2<90^\circ\implies\sin\dfrac A2>0$

For the other part,

$$y=1-2\sin^2\frac A2+2\sin\frac A2\cos\frac{B-C}2$$

$$\iff2\sin^2\frac A2-\sin\frac A2\cdot2\cos\frac{B-C}2+y-1=0$$ which is a Quadratic equation in $\sin\dfrac A2$ which is real

So, the discriminant must be $\ge0$

i.e., $\left(2\cos\dfrac{B-C}2\right)^2-4\cdot2(y-1)\ge0$

$\iff 4y\le4+2\cos^2\dfrac{B-C}2=4+1+\cos(B-C)\le4+1+1$

The equality occurs iff $\cos(B-C)=1\iff B=C$ as $0<B,C<180^\circ$

where $\sin\dfrac A2=\dfrac12\implies\dfrac A2=30^\circ$ as $0<\dfrac A2<90^\circ$

Solution 2:

better is to show that $\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$ where $r$ is the inradius and $R$ is circumradius of the given triangle. Your theorem follows from this equation